GCDs and Gauss' Lemma R. C. Daileda 1 GCD Domains Let R be a domain and S ⊂ R. We say c 2 R is a common divisor of S if cjs for every s 2 S. Equivalently, S ⊂ (c) or (S) ⊂ (c). We say a common divisor d is a greatest common divisor (GCD) of S if every common divisor c of S satisfies cjd. That is, d is a GCD of S if and only if (d) is the least element of the set f(c) j c 2 R; (S) ⊂ (c)g; (1) provided the least element exists. When it exists, the GCD of S is only defined up to association: the GCDs of S are the generators of the least element of (1). We will write a ≈ b to indicate that a; b 2 R are associates. In this case their common equivalence class is aR×. The association classes form a commutative multiplicative monoid under the operation (aR×)(bR×) = abR× (see the appendix). Let gcd S denote the association class consisting of the GCDs of S. Thus, the statement \d is a GCD of S" is equivalent to d 2 gcd S. A domain R is called a GCD domain if every finite subset of R has a GCD. Suppose that R is a GCD domain and d is a GCD for a1; a2; : : : ; am 2 R. Write ai = dbi for each i, with bi 2 R. If c 2 R and cjbi for all i, then cdjai for all i. This means cdjd and hence cj1, so that c is a unit. It × follows that 1 is a GCD for the ai and hence gcdfaig = R . Example 1. a. In Z, gcd(49; 21) = {±7g. In Z[X], gcd(2;X) = {±1g. Notice that in the first case we have (49; 21) = (7), while in the second (2;X) 6= (1) = Z[X]. b. Every PID R is a GCD domain. Given any S ⊂ R,(S) = (d) for some d 2 R, and hence gcd S = dR×. c. Every B´ezoutdomain R is a GCD domain for similar reasons. d. Every UFD is a GCD domain. See section 3. + e. Let F be a field. The monoid ring F [X; Q0 ] is an GCD domain but is not a UFD. f. Every GCD domain is an AP domain,. See Theorem 1. Remark 1. a. There are now two different notions of the GCD of S: (i) the the smallest ideal containing S; (ii) (the generators of) the smallest principal ideal containing S. These need not be the same! Take R = Z[X] and S = f2;Xg. The ideal-theoretic GCD is the proper ideal (2;X) = ff 2 Z[X] j f(0) eveng, while the element-wise GCDs are {±1g. The reason for the distinction here (and in general) is that S does not generate a principal ideal. b. Consider R as a subring of its quotient field. If a; b 2 R, b 6= 0 and bja, then there is a unique c 2 R so that bc = a. We call c the factor or divisor complementary to b. We can use fractions to help us bc a c a represent it. If we embed R in its quotient field, bc = a becomes 1 = 1 , which is equivalent to 1 = b . a When bja in R we will therefore write b to denote the divisor complementary to b. Lemma 1. Let R be a domain and S; T ⊂ R. 1 gcd S 1. If c 2 R is nonzero, S ⊂ (c) and gcd S exists, then gcd(S=c) = . c 2. For any c in R, gcd(cS) = c · gcd S, provided gcd(cS) exists. 3. gcd(S [ T ) = gcd(S [ gcd T ), provided the GCDs exist. d Proof. 1. Let d 2 gcd S. Then cjd by definition. So S ⊂ (d) implies S=c ⊂ c . If S=c ⊂ (e), then S ⊂ (ce) d d × gcd S so that (d) ⊂ (ce) and hence c ⊂ (e). This proves that gcd(S=c) = c R = c . 2. If c = 0 there is nothing to prove. Otherwise, if gcd(cS) exists, then since cS ⊂ (c), the first part implies cS gcd(cS) gcd S = gcd = ) c · gcd S = gcd(cS): c c 3. For any d 2 R, T ⊂ (d) if and only if gcd T ⊂ (d). So S [ T ⊂ (d) , S ⊂ (d) and T ⊂ (d) , S ⊂ (d) and gcd T ⊂ (d) , S [ gcd T ⊂ (d): The result follows. Lemma 2. Let R be a GCD domain, let c 2 R and let S; T ⊂ R be finite. Then gcd(S [ cT ) = gcd(S [ c · gcd(S [ T )): Proof. Repeatedly apply parts 2 and 3 of Lemma 1: gcd(S [ cT ) = gcd(gcd(S) [ cT ) = gcd(gcd(S [ cS) [ cT ) = gcd(S [ cS [ gcd(cT )) = gcd(S [ cS [ c · gcd(T )) = gcd(S [ gcd(cS [ c · gcd(T ))) = gcd(S [ c · gcd(S [ gcd(T ))) = gcd(S [ c · gcd(S [ T )): Corollary 1. Let R be a GCD domain, a; b; c 2 R. If gcd(a; b) = R×, then gcd(a; bc) = gcd(a; c). Proof. Apply Lemma 2 with S = fag and T = fbg. Corollary 2. Let R be a GCD domain, a; b; c 2 R. If gcd(a; b) = gcd(a; c) = R×, then gcd(a; bc) = R×. Under the stronger hypothesis that R is a B´ezoutdomain, Corollaries 1 and 2 have dramatically simpler proofs. The examples of GCDs domains that we have seen so far are all AP domains. Corollary 2 can be used to show that there's a reason for this. Theorem 1. Every GCD domain is an AP domain. Proof. Let R be a GCD domain and suppose a 2 R is irreducible. Suppose b; c 2 R and ajbc. Then gcd(a; bc) = aR× 6= R×. By Corollary 2, gcd(a; b) 6= R× or gcd(a; c) 6= R×. Without loss of generality assume that gcd(a; b) 6= R×. Since the only divisors of a are its associates and units, it must be the case that gcd(a; b) = aR×. In particular, ajb, and a is prime. 2 2 Gauss' Lemma P i Let R be a GCD domain. Given a polynomial f(X) = i aiX 2 R[X], we define the content of f to be c(f) = gcd(a0; a1; a2;:::): × The GCD exists because only finitely many of the ai are nonzero. When c(f) = R say that f is primitive. As we will see, in terms of factorization, the primitive polynomials over a domain play the same role the monic polynomials over a field play. If 0 6= d 2 c(f), then X ai f = d Xi = df;e d i where fe is primitive by part 1 of Lemma 1. Furthermore, by part 2 of Lemma 1 we have c(ef) = e · c(f) for any r 2 R. In particular, if a 2 R, then c(a) = a · c(1) = aR×. This means that a 2 R is primitive if and only if a 2 R×. These are the only observations we need to prove Gauss' lemma. Lemma 3 (Gauss). Let R be a GCD domain and let f; g 2 R[X]. If f and g are primitive, then fg is primitive. Proof. We prove the contrapositive by induction on n = deg fg. When n = 0, f; g 2 R and c(fg) = fgR× = (fR×)(gR×) = c(f)c(g). Since c(fg) 6= R×, this implies c(f) 6= R× or c(g) 6= R×. Now let n ≥ 1 and assume we have proven the result for all pairs of polynomials whose product has degree less than n. Suppose deg fg = n. Write c(fg) = dR× 6= R×. Let f = aX` + O(X`−1) and g = bXm + O(Xm−1). Then fg = abXn +O(Xn−1) and ab 2 (d). Thus (ab; d) = (d) 6= R and either gcd(a; d) 6= R× or gcd(b; d) 6= R×, by Lemma 2. Assume, without loss of generality, that gcd(a; d) = eR× 6= R×. Then e divides every coefficient of fg −aXmg = (f −aX`)g. This implies that c((f −aX`)g) ⊂ (e) 6= R and hence (f −aX`)g is imprimitive. Since deg(f − aX`) < deg f, deg(f − aX`)g < deg fg = n. The inductive hypothesis therefore implies that ` either f1 = f − aX or g is imprimitive. × If g is imprimitive, we're finished. If g is primitive, write f1 = e1fe1 where c(f1) = e1R and fe1 2 R[X] is primitive. Because deg fe1 = deg f1, we can again apply the (contrapositive of the) inductive hypothesis to conclude that fe1g is primitive. Then × c(f1g) = c e1fe1g = e1c fe1g = e1R = c(f1): ` It follows that c(f1) ⊂ (e). So e divides all the coefficients of f1 as well as the coefficients of aX . Thus e m × divides all of the coefficients of f1 + aX = f. That is, c(f) ⊂ (e). Since (e) 6= R, c(f) 6= R so that f is imprimitive. This completes the induction. Corollary 3. Let R be a GCD domain and let f; g 2 R[X]. Then c(fg) = c(f)c(g). Proof. If f = 0 or g = 0 there is nothing to prove, so we may assume f; g 6= 0. Then f = dfe and g = ege, where d 2 c(f), e 2 c(g) and f;e ge 2 R[X] are primitive. Then fege is primitive by Gauss' lemma so that × c(fg) = c defege = de · c fege = deR = c(f)c(g): There is a somewhat simpler and more intuitive proof of Gauss' lemma when R is a a UFD.